250 Problems in Elementary Number Theory - Sierpinski (1970)

SOLUTIONS 2510. These are all odd numbers > 1. In fact, if n is odd and > 1, thenthe number (n-1)/2 is a positive integer, and for k = 1, 2, ... , (n-l)J2 weeasily getthus n\I"+2"+ ... +(n-l)".On the other hand, if n is even, let 2 Sbe the highest power of 2 whichdivides n (thus, s is a positive integer). Since 2 S ~ s, for even k we have2 S \k", and for odd k (the number of such k's in the sequence 1, 2, ... , n-l1-2 s - 1 (is in) we have, by Euler's theorem, I{,- = 1 mod 2~, hence k" = 1 (mod 2~(since 2 S 1- !n). Thereforewhich impliesin view of the fact that 2"+4"+ ... + (n-2)" == 0 (mod 2~. Now, if we hadn\I"+2"+ ... +(n-1)", then using the relation 2 s ln we would have in= 0 (mod 2~, hence 2 s lln and 2 S + 1 In, contrary to the definition of s. Thus,for even n we have n,r 1"+2"+ ... +(n-l)".REMARK. It follows easily from Fermat's theorem that if n is a prime,then nll"-1+2"-I+ ... +(n-l),,-I+ 1; we do not know any composite numbersatisfying this relation. G. Ginga conjectured that there is no such compositenumber and proved that there is no such composite number n < 10 1000 •11. Consider four cases:(a) n = 4k, where k is a positive integer. Thena" = 28IC+1_24k+1+1 = 2-2+1 = 1 (mod 5),b" = 28k+l+24k+l+1 == 2+2+1 = 0 (mod 5)(since 24 = 1 (mod 5), which implies 2 4k · 2 8k = 1 (mod 5»).(b) n = 4k + 1, k = 0, 1, 2, .... Thena" = 28k+3_24k+2+1 = 8-4+1 = 0 (mod 5),h" = 28k+3+24k+2+1 == 8+4+1 = 3 (mod 5).